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To show how to solve the equation, create two functions and find their intersection points. What two functions can be used to solve the following equation, and what is their solution set?

10−x^2=x+4

(1 point)
Responses

g(x)=x+4
, f(x)=10+x^2
, {−3,2}
g left parenthesis x right parenthesis equals x plus 4 , f left parenthesis x right parenthesis equals 10 plus x squared , left brace negative 3 comma 2 right brace

f(x)=10−x^2
, g(x)=x+4
, {−3,2}
f left parenthesis x right parenthesis equals 10 minus x squared , g left parenthesis x right parenthesis equals x plus 4 , left brace negative 3 comma 2 right brace

f(x)=10+x^2
, g(x)=x+4
, {−3,1}
f left parenthesis x right parenthesis equals 10 plus x squared , g left parenthesis x right parenthesis equals x plus 4 , left brace negative 3 comma 1 right brace

g(x)=x+4
, f(x)=10−x^2
, {1,6}

1 answer

To solve the equation \(10 - x^2 = x + 4\), we can create two functions based on it:

  1. Let \(f(x) = 10 - x^2\)
  2. Let \(g(x) = x + 4\)

Now, we want to find the intersection points of these two functions, which means we are looking for the values of \(x\) where \(f(x) = g(x)\).

To find the intersection points, set the two functions equal to each other:

\[ 10 - x^2 = x + 4 \]

Now, rearranging the equation gives:

\[ 10 - 4 = x + x^2 \ 6 = x + x^2 \ x^2 + x - 6 = 0 \]

Now we can factor this quadratic equation:

\[ (x - 2)(x + 3) = 0 \]

Setting each factor to zero gives the solutions:

\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]

Thus, the intersection points are \(x = -3\) and \(x = 2\).

Therefore, the solution set is:

\[ {-3, 2} \]

In conclusion, the two functions can be written as:

  • \(f(x) = 10 - x^2\)
  • \(g(x) = x + 4\)

The solution set is: \({-3, 2}\)

So the correct answer from the choices provided is:

f(x)=10−x^2, g(x)=x+4, {−3,2}